Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define ll long long

ll mod=(ll)1e18+9, bas=1e9+7, inv;

ll mul(ll x,ll y){
    ll t=0;
    for (;y;y>>=3,x=x*8%mod)
        t=(t+x*(y&7))%mod;
    return t;
}

ll K(ll x,ll y=mod-2){
    if (y<0) return K(K(x),-y);
    ll t=1;
    for (;y;y>>=1,x=mul(x,x))
        if (y&1) t=mul(t,x);
    return t;
}

ll n, ans, p[505000], f[505000];
char s[505000];
map<ll,ll>mp;

int main(){
    cin>>n>>s+1;
    p[0]=1; inv=K(bas);
    for (int i=1;i<=n;++i){
        p[i]=p[i-1]; f[i]=f[i-1];
        if (s[i]=='<') p[i]=mul(p[i],inv);
        if (s[i]=='>') p[i]=mul(p[i],bas);
        if (s[i]=='+') f[i]=(f[i]+p[i])%mod;
        if (s[i]=='-') f[i]=(f[i]-p[i]+mod)%mod;
        ++mp[f[i]];
    }
    for (int i=0;i<n;++i){
        ans+=mp[(mul(f[n],p[i]) + f[i])%mod];
        --mp[f[i+1]];
    }
    cout<<ans<<endl;
}